video. If a Surface $S$ is compact and connected, if the Gaussian curvature $K$ is a constant. Then, $S$ is a Sphere!

In fact because of compactness, constant $K$, implies that $K>0$. First, because any compact surface has an Elliptic point (see argument using Height function, and existance of global maximum and minimum for Compact spaces), $K\geq 0$. Furthermore, it cannot be $0$ everywhere because that would imply that every point belongs to some path along which the normal is constant (following locally the Principal direction with eigenvalue $0$, necessary for $K=0$). Note that compactness, which implies existence of maximum and minimum in particular for the Height function, then implies the existence of a point in the surface with any normal vector. Hmm.. However, we can parametrized the surface by following the curves with constant normal vector, and a curve that crosses all of them. This doesn't leave enough space for all normal vectors to be present in the surface.. Not very formal, but I think it works.

$K>0$ implies the Mean curvature $H$ is non-0, then the rest of the argument follows as in Jellet-Liebmann theorem